QJEEM21Ma4c8d: JEE Main 2021 Mathematics - Class 11 JEEM21Ma4c8d Solution: If the variance of 10 natural numbers 1, 1, 1, ....., 1, k is…

Question

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If the variance of 10 natural numbers 1, 1, 1, ....., 1, k is less than 10, then the maximum possible value of k is ________.

Accepted Answer

{\sigma ^2} = {{\sum {{x^2}} } \over n} - {\left( {{{\sum x } \over n}} \right)^2}

{\sigma ^2} = {{(9 + {k^2})} \over {10}} - {\left( {{{9 + k} \over {10}}} \right)^2} &lt; 10

(90 + {k^2})10 - (81 + {k^2} + 8k) &lt; 1000

90 + 10{k^2} - {k^2} - 18k - 81 &lt; 1000

9{k^2} - 18k + 9 &lt; 1000

{(k - 1)^2} &lt; {{1000} \over 9} \Rightarrow k - 1 &lt; {{10\sqrt {10} } \over 3}

k &lt; {{10\sqrt {10} } \over 3} + 1

k

\le

11

Maximum integral value of k = 11.

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